Solutions

Solutions, Colloids, and SuspensionsSolution Formation and Factors Affecting Solubility

Solubility CurvesMolarity and Solubility Product Constant (Ksp)

Colligative PropertiesKey

In Your TextbookSolutions, Suspensions, and Colloids: pp. 482–83 and pp. 490-493

Factors Affecting Solubility: pp. 501 - 508Solubility Curves: pp. 504 – 505

Molarity and Ksp: pp. 509–512 and pp. 630-637Colligative Properties: pp. 517 - 524

Assignment 1: Defining Solutions

Compare and contrast the contents of the three test tubes. Record observations in the chart below.

A B C

Green clearliquid

Whitecloudyliquid

Browncloudy liquidwith particlesat the bottom.

Read pp. 482-483, pp. 490-494, pp. 502-508, and pg354 in your textbook.

Which of the three test tubes contains a solution. Explain why.The first tube has a solution because the particles are dissolved as small as they can get. It is not at all cloudy.

How could I test to see if you are correct?I could look for a Tyndall effect by shining a light through the mixture andsee if the light is scattered (you can see the beam of light in the mixture as it passes through). If you can, it is not a solution.

Which of the following would also be a solution?Blood no (colloid)Mayonnaise no (colloid)Italian Salad Dressing no (suspension)Coca-Cola yesAir yesOcean Water no/yes The water itself would be a solution, but if you scooped up a sample, it would have all kinds of things in it (a suspension).

Define: Solubility, Solute, SolventSolvent: dissolves the soluteSolute: is dissolved by the solventSolubility: the amount of solute that will dissolve in a given amount of water at a certain temperature.

Tonight, find 10 things in your home that are classified as solutions. Make a list of these solutions. (What is the solute and the solvent in each solution?)Some examples: windex, coke, Listerine, vinegar…..

Assignment 2: Solutions and Molarity Review.

Solubility: Predict which of the following will dissolve in water and why. Look up the structures and consider the bonding, structure, and the intermolecular forces.

Sodium chloride (table salt) NaCl is ionic so is likely to dissolve.

Sucrose (sugar) C12H22O11 is covalent, but is polar so will dissolve to some degree. These molecules have dipole-dipole interactions.

Vegetable oilCH3(CH2)16COOH is a long carbon chain and they tend to be nonpolar so not soluble. It would have only dispersion forces.

Isopropyl alcohol (rubbing alcohol)C4H9OH is polar covalent and would dissolve in water. It has dipole-dipole interactions, specifically H-bonding.

Acetic Acid (vinegar)HC2H3O2 is polar covalent (actually ionizes as an acid in water) so would dissolve. It would have dipole-dipole attractions.

What does the rule “like dissolves like” state? This rule says that polar solvents dissolve solutes that are polar or ionic (have a + and a -) and solvents that are non-polar dissolve solutes that are non-polar (turpentine dissolves non-polar solutes).

Test your hypothesis by dissolving a small amount of the substance in water. Are there any that do not follow the rule “like dissolves like”? Are any of these substance already in solution when you buy them from the grocery store?These all follow the rule. The alcohol and vinegar are already in water when you buy them.

Molarity Review:

1. A salt solution has a volume of 250 mL and contains 0.70 mol of sodium chloride. What is the molarity of the solution?

0.70 moles NaCl = 2.8 M 0.25 L

2. A solution of glucose has a volume of 2.0 L and contains 36.0 g of solute. What is the molarity of the solution?

36.0 g C6H12O6 x 1 mole = 0.20 moles 180.16 g

M = 0.20 moles = 0.10 M 2.0 L solution

3. How many moles of solute are in 250 mL of 2.0 M calcium chloride solution? How many grams would that be?

2.0 M = x moles x = 0.50 moles Ca Cl2

0.25 L

0.50 moles CaCl2 x 110.98 g CaCl2 = 55 g CaCl2

1 mole CaCl2

4. Describe how you would make a 0.25 M solution of calcium hydroxide. (Use sketch to illustrate.)

Put 0.25 moles (19 grams) of Ca(OH)2 in a volumetric flask and add water to the 1 L mark.

5. What is the molarity of the sugar solution in a can of coke? (Assume sugar to be sucrose)

39 g C12H22O11 x 1 mole = 0.11 mole M = 0.11 mole = 0.31M 342 g .355 L

Assignment 3: Factors Affecting Solubility:

Effect Reason Does SolubilityChange?

TemperatureChange of

Solids

TempertureChange of

Gases

Stirring

Particle Size

PressureOf

Gases

Assignment 4: Ksp Problems:1. Write the equation for the dissolving and the solubility product constant

expression for the following solids:

NaCl Ksp = [Na] [Cl]

Ag2SO4 Ksp = [Ag]2 [SO4]

Pb(OH)2 Ksp = [Pb] [OH]2

Al2(SO4)3 Ksp = [Al]2 [SO4]3

2. The solubility of PbSO4 in water is 0.038 grams per liter. Calculate the solubility product constant.

PbSO4 (s) Pb+2 (aq) + SO4-2 (aq) Ksp = [Pb+2][SO4

-2]

0.038g x 1 mole = 1.25 x 10-4 moles in 1 liter 303 g

Ksp = [1.25 x 10-4] [1.25 x 10-4] = 1.6 x 10-8

3. The solubility of silver(I) chromate in water is 0.024 g per liter. Determine the solubility product constant.

Ag2CrO4 (s) 2 Ag+1 (aq) + CrO4-2 (aq) Ksp = [Ag+1]2 [CrO4

-2]

0.024g x 1 mole = 7.2 x 10-5 moles in 1 liter 332 g

Ksp = [2(7.2 x 10-5)]2 [7.2 x 10-5] = 1.0 x 10-8

4. If the Ksp of barium chromate is 8.3 x 10-11, calculate the solubility in moles per liter.

BaCrO4 (s) Ba+2 (aq) + CrO4-2 (aq) Ksp = [Ba+2][CrO4

-2]I x 0 0

C -x +x +xE 0 x x

8.3 x 10-11 = x2

x = 9.2 x 10-6 M 5. If the Ksp for strontium chromate is 3.6 x 10-5, would a solution in which 1.02

grams is dissolved in one liter be saturated?SrCrO4 (s) Sr+2 (aq) + CrO4

-2 (aq) Ksp = [Sr+2][CrO4-2]

I x 0 0 C -x +x +x

E 0 x x

3.6 x 10-5 = x2 0.0060 moles x 203.6 g = 1.2 g will dissolve in x = 0.0060 M 1 mole 1 liter (unsaturated)

6. What happenes if the product of the concentrations of A+ ion and B- ion is greater than the Ksp value?The solution would be supersaturated and would not be stable. It would precipitate is shaken or given a seed crystal.

7. Will calcium carbonate precipitate if the Ca+2 concentration in a solution is 0.0002M and the CO-2 concentration is the same. The Ksp is 4.8 x 10-9.

CaCO3 (s) Ca+2 (aq) + CO3-2 (aq) Ksp = [Ca+2][CO3

-2]I x 0 0

C -x +x +xE 0 x x

4.8 x 10-9 = x2 yes, 0.0002 is larger than 0.000060x = 6.0 x 10-5 M which is a saturated solution.

8. Seawater is saturated with AgCl. The [Cl-] in seawater is 0.53M and the Ksp for AgCl is 1.8 x 10-10. Calculate the [Ag+]

AgCl (s) Ag+1 (aq) + Cl-1 (aq) Ksp = [Ag+1][Cl-1]

1.8 x 10-10 = [Ag+1] [0.53] *There are lots ofx = 3.4 x 10-10 M salts in seawater, so

not all Cl-1 is from AgCl. You shift thereaction

9. When 12 M HCl is added to a saturated solution of NaCl, a precipitate forms. Explain why?

NaCl (s) Na+1 (aq) + Cl-1 (aq) Ksp = [Na+1][Cl-1]

By adding Cl-1 ions, the reaction is shifted toward the reactants (the solid) so a precipitate forms.

Challenge:

The Ksp for PbF2 is 3.7 x 10-8. Calculate the grams of each ion dissolved on one liter of a saturated solution.

PbF2 (s) Pb+2 (aq) + 2 F-1 (aq) Ksp = [Pb+2][F-1]2

I x 0 0 C -x +x +2x

E 0 x 2x

3.7 x 10-8 = (x)(2x)2

3.7 x 10-8 = 4x3

x = 2.1 x 10-3 moles Pb+2 in one liter and 2(2.1 x 10-3) molesof F-1 in one liter.

2.1 x 10-3 x 1 mole PbF2 = 0.51 grams PbF2 in 1 liter 245 g

Assignment 5: Solubility Curves

1. Does heat increase the solubility of all compounds the same amount? no

2. How many grams of solute are needed to saturate 100.0 g of water at 80.0oC

to prepare a saturated solution of:

Potassium bromide 98g

Sodium chloride 40g

3. At what temperature are sodium chloride and potassium nitrate equally soluble? At about 22oC

4. Is sodium chloride of potassium nitrate more soluble at 20.0oC? NaCl

5. What is the molality of a saturated solution of potassium bromide at 50.0oC?

80g of KBr will dissolve in 100g of water at 50oC

80g x 1 mole = 0.67 moles m = 0.67 moles KBr = 6.7m119 g KBr 0.1 kg water

6. What is the molality of a saturated solution of sodium chlorate at 10.0oC?

95g of NaClO3 will dissolve in 100g of water at 10oC

95g x 1 mole = 0.89 moles m = 0.89 moles NaClO3 = 8.9m 106.5 g NaClO3 0.1 kg water

7. Is a 2.0M solution of sodium chloride possible at 30.0oC?2M = x moles = 0.2 moles NaCl

0.1 L

0.20 moles NaCl x 58.5 g NaCl = 11.7 grams NaCl in 0.1 L water (about 0.1kg or 100g)

The chart indicates that 40g NaCl will dissolve in 100g H2O at 30.0oC, so it is possible to make a 2.0 M solution.

8. Are the following solutions saturated, unsaturated, or supersaturates?

40.0g of potassium bromide in 100 g of water at 20.0oC

The chart indicates that 70g of KBr will dissolve in 100g water at 20oC, so 40g in 100g water would be unsaturated.

40.0g of potassium bromide in 50.0g of water at 50.0oC

The chart indicates that 80g of KBr will dissolve in 100g of water at 50oC, so half that amount would dissolve in 50g water:

80g KBr = x g KBr x = 40g 100g H2O 50g H2O

The solution would be saturated.

20.0g of sodium chloride in 50.0g of water at 10.0oC

The chart indicates that 35g of NaCl will dissolve in 100g water at 10oC, so half that amount will dissolve in 50g water.

35g NaCl = x g NaCl x = 17.5g 100g H2O 50g H2O

So, if 20.0g of NaCl was dissolved in 50g of water, it would be supersaturated.

Book Problems:

25. PbS (s) Pb+2 (aq) + S-2 (aq) Ksp = [Pb+2][S-2]I x 0 0

C -x +x +xE 0 x x

3 x 10-28 = x2

x = 2 x 10-14 M PbS is a saturated solution

27. Ag2S (s) 2 Ag+1 (aq) + S-2 (aq) Ksp = [Ag+1]2 [S-2]I x 0 0

C -x +2x +xE 0 2x x

8 x 10-51 = (2x)2(x)8 x 10-51 = 4x3 x = 1 x 10-17 so, [Ag+] = 2 (1 x 10-17) = 2 x 10-17 M

[S-2] = 1 x 10-17 M(or M 1 x 10-17 Ag2S is a saturated solution)

29. FeS (s) Fe+2 (aq) + S-2 (aq) Ksp = [Fe+2][S-2] x 0.04 x

-x +x +x 0 0.04 + x x

*Because the Ksp is so small, you know the x in 0.04 – x above will be negligible in comparison to 0.04, so you can leave the x out.

8 x 10-19 = [0.04] xx = [2 x 10-17]

31. Ca(NO3)2 (s) Ca+2 (aq) + 2 NO3-1 (aq)

0.001 0 0 *This solution would -0.001 +0.001 +0.002 provide 0.001M Ca-2

0 0.001 0.002

Na2CO3 (s) 2 Na+1 (aq) + CO3-1 (aq)

0.0008 0 0 *This solution would -0.0008 +0.0016 +0.0008 provide 0 0.0016 0.0008 0.0008M CO3

-2

so:

CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)

0.001 0.0008

Ksp = [0.001] [0.0008] = 8 x 10-7

This number is higher than the actual Ksp of 4.5 x 10-9 so a ppt would form!

Assignment 6: More Solubility Curves

1. Most substances on this graph show increased solubility as the temperature increases. What are the exceptions?

NH3 (gas) and Ce2(SO4)3

2. Each curve shows how solubility for that substance changes as temperature

changes. The solubilities of substances whose curves show greater (steeper)

slopes are mores (more/less) affected by temperature changes than those that have

more gradual slopes.

NaCl 3. Which salt has solubility values that are least affected by changes in temperature?

96oC 4. At what temperature do potassium chlorate and potassium chloride have the same solubility in water?

KClO3 5. Which compound is least soluble in water at 12oC?

KNO3 6. A saturated solution of which compound contains 130 grams of solute per 100 grams of water at 70oC?

40 g 7. How many grams of sodium chloride are required to saturate 100 grams of water at 100oC?

200g 8. How many grams of sodium chloride are required to saturate 500 grams of water at 100oC? 40g NaCl = x g NaCl x = 200g NaCl

100g H2O 500g H2O160g 9. How many grams of sodium nitrate are required to saturate 200 grams of water

at 10oC? 80g NaNO3 = x g NaNO3 = 160g NaNO3

100g H2O 200g H2O13g 10. At 50oC, 100 grams of water is saturated with cerium(III) sulfate. How many

grams of cerium(III) sulfate must be added to saturate the solution if the temperature is changed to 0oC? 5g Ce2(SO4)3 at 50oC, 18g Ce2(SO4)3 at 0oC

18g – 5g = 13g need to be added20g 11. At 50oC, 100 grams of water is saturated with potassium nitrate. How many

grams of potassium nitrate will precipitate when the solution is cooled to 40oC? 82g KNO3 at 50oC, 62g KNO3 at 40oC

82g – 62g = 20g will fall out.13% 12. What is the percent by mass of a saturated solution of potassium chlorate at

40oC? 15g KNO3 x 100 = 13%115g total

10.8m 13. What is the molarity of a saturated solution of NaNO3 at 25oC? 92g NaNO3 x 1 mole NaNO3 = 1.08 moles

85.0g NaNO3

m = 1.08 moles NaNO3 = 10.8m 0.100 kg H2O

117 14. What is the Ksp of the Na NO3 at 25oC?NaNO3 (s) Na+1 (aq) + NO3

-1 (s)

Ksp = [Na+1] [NO3-1]

[10.8] [10.8] = 117

Assignment 7: Yet More Solubility Curves

20g 1. At 80oC, 100 grams of water is saturated with potassium chloride. How many grams of solute will precipitate when the solution is cooled to 45oC

82g at 50oC, 62g at 40oC 82g – 62g = 20g ppt.5g 2. If 50 grams of water saturated with potassium chlorate at 23oC is slowly

evaporated to dryness, how many grams of salt will be recovered? 10gKClO3 = xg KClO3 x = 5g 100g H2O 50g H2O

20g 3. If 30 grams of potassium chloride is dissolved in 100 grams of water, how many additional grams would be needed to make the solution saturated at

80oC?50g will go in 100g H2O at 80oC; 50g – 30g = 20g

51g 4. What is the smallest mass of water required to dissolve completely 23 grams of ammonium chloride at 40oC?

45g NH4Cl = 23g NH4Cl x = 51g H2O100g H2O x g H2O

1g/oC 5. A saturated solution of sodium nitrate in 100 grams of water at 40oC is heated to 50oC. What is the rate of increase in solubility in grams per degree??

105g NaNO3 at 40oC115g NaNO3 at 50oC

NH4Cl 6. Which chloride has the greatest percent by mass at 60oC? What is the percent? 55g NH4Cl x 100 = 35% 155g total

0.18M 7. What is the molarity (molality) of a saturated solution of cerium(III) sulfate at 25oC?

10g Ce2(SO4)3 x 1 mole Ce2(SO4)3 = 0.018 moles 568g Ce2(SO4)3

M = 0.018 moles Ce2(SO4)3 = 0.18M0.100 L solution

0.020 8. What is the Ksp of a saturated solution of cerium(III) sulfate at 25oC?Ce2(SO4)3 (s) 2 Ce+3 (aq) + 3 SO4

-2 (aq) [0.18] 0 0 - 0.18 + 0.36 + 0.54 0 [0.36] [0.54]

Ksp = [Ce+3]2 [SO4-2]3 = [0.36]2 [0.54]3 = 0.020

-30oC 9. A solution of KNO3 is saturated at 50oC. At what temperature would the solution freeze 82g KNO3 dissolve in 100g H2O at 50oC.

82g KNO3 x 1 mole KNO3 = 0.81 moles m = 0.81 moles = 8.1m 101g KNO3 0.100 kg H2O

KNO3 (s) K+1 (aq) + NO3-1 (aq)

[8.1] 0 0-8.1 +8.1 +8.1

0 [8.1] [8.1] = 16.2 m total particles

tf = 1.86oC/m x 16.2m = - 30oC3.3m 10. A solution of KNO3 freezes at -12oC. At what temperature would the solution

be saturated? -12oC = 1.86oC/m x m m = 6.5m total particles

Because KNO3 breaks into 2 particles and the total is 6.5, it must be half that for each ion or 3.3m each

KNO3 (s) K+1 (aq) + NO3-1 (aq)

[3.3] 0 0 -3.3 +3.3 +3.3

0 [3.3] [3.3] = 6.5m total

Assignment 8: Molarity, molality, and colligative properties

1. What is the molarity of a solution in which 82.0g of calcium nitrate is dissolved in

enough water to make 500.0 mL of solution?82.0g Ca(NO3)2 x 1 mole = 0.4998 moles

164.09g

M = 0.4998 mol = 1.00M 0.5000 L soln

2. What is the molality of a solution in which 50.0g of copper(II) sulfate is dissolved

in 250.0mL of water50.0g CuSO4 x 1 mole CuSO4 = 0.313 moles

159.5 g CuSO4

m = 0.313 moles CuSO4 = 1.25 m 0.2500 kg H2O

3. Calculate the mass of solute in 250.0mL of sodium sulfate solution that is 2.00M

(molar).2.00M = x moles Na2SO4 = 0.500 moles

0.2500 L soln

0.500 moles Na2SO4 x 142.04g Na2SO4 = 71.0g Na2SO4 1 moles Na2SO4

4. Calculate the mass of solute in 250.0mL of sodium sulfate solution that is 2.00m

(molal).2.00M = x moles Na2SO4 = 0.500 moles ~ 0.2500 kg H2O

0.500 moles Na2SO4 x 142.04g Na2SO4 = 71.0g Na2SO4 1 moles Na2SO4

5. The Ksp of silver iodide is 8.3 x 10-17 at 25oC; what is the molarity of a saturated

solution? Write the reaction for the dissolving.

AgI (s) Ag+1 (aq) + I-1 (aq) x 0 0 - x + x + x 0 x x

Ksp = [Ag+1] [I-1] 8.3 x 10-17 = x2 x = 9.1 x 10-9 M

6. A saturated solution of lead(II) chloride at 25oC contains 2.2 grams of solute in

500.0 mL. What is the Ksp? Write the reaction for the dissolving.

2.2g PbCl2 x 1 mole PbCl2 = 7.9 x 10-3 moles278.1g PbCl2

M = 7.9 x 10 -3 moles = 0.016 M0.5000 L

PbCl2 (s) Pb+2 (aq) + 2 Cl-1 (aq) [0.016] 0 0 - 0.016 + 0.016 + 0.032 0 [0.016] [0.032]

Ksp = [Pb+2] [Cl-1]2 = [0.016] [0.032]2 = 1.6 x 10-5

Is the lead(II) chloride or the silver iodide in the previous problem more soluble?

The lead(II) chloride is more soluble as the Ksp is higher (1.6 x 10-5 vs 8.3 x 10-17)

7. What is the freezing point of a 0.85 molal solution of sugar? Kf = 1.86oC/m.

C12H22O11 (s) C12H22O11 (aq) 0.85 0 tf = 1.86oC/m x 0.85m = 1.6oC - 0.85 + 0.85 so the freezing pt. is – 1.6oC 0 0.85

8. What is the freezing point of a solution that contains 68.5 grams of sucrose, C12H22O11, dissolved in 100.0 grams of water?

68.5g C12H22O11 x 1 mole = 0.200 moles 342g

m = 0.200 moles = 2.00 m0.1000 kg

C12H22O11 (s) C12H22O11 (aq) 2.00 0 tf = 1.86oC/m x 2.00m = 3.72oC - 2.00 + 2.00 so the freezing pt. is – 3.72oC 0 2.00

9. What is the freezing point of a solution that contains 68.5 grams of salt, NaCl, dissolved in 100.0 grams of water?

68.5g NaCl x 1 mole = 1.17 moles 58.5g

m = 1.17 moles = 11.7 m0.1000 kg

NaCl (s) Na+1 (aq) + Cl-1 (aq) 11.7 0 0 tf = 1.86oC/m x 23.4m = 43.5oC - 11.7 + 11.7 +11.7 so the freezing pt. is – 43.5oC 0 11.7m 11.7m

23.5m total

Why is CaCl2 used on the roads instead of NaCl or sucrose?

When CaCl2 dissolves, 3 particles are produces. When NaCl dissolves, only 2 particles are produced.

10. What is the boiling point of the solution in problem #9?

68.5g NaCl x 1 mole = 1.17 moles 58.5g

m = 1.17 moles = 11.7 m 0.1000 kg

NaCl (s) Na+1 (aq) + Cl-1 (aq)11. 0 0 tb = 0.512oC/m x 23.5m

- 11.7 + 11.7 +11.7 tb = 12oC 0 11.7m 11.7m so the boiling pt. is 112oC

23.5m total

Assignment 10: Solutions ReviewDefine:1. Solution2. Solvent3. Solute4. Homogeneous5. Heterogeneous6. Soluble7. Insoluble8. Solvation9. Colligative property10. Saturated solution (how do you tell?)11. Unsaturated solution (how do you tell?)12. Supersaturated solution (how do you tell?)Questions:Write the reaction for the dissolving of sucrose in water. Does sucrose dissociate?

C12H22O11 (s) ------- C12H22O11 (aq) Does not dissociate. Molecules stay together.

Write the reaction for the dissolving of potassium bromide in water? Does KBr dissociate?

KBr (s) ------ K+ (aq) + Br- (aq) Does dissociate into ions.

Write the Ksp expression for the above equations. What does Ksp describe? What does a large Ksp mean?

Ksp = [K+ ] [Br- ] Ksp = [C12H22O11]The solubility product constant describes how far the reaction goes toward the products which is the dissolved for of the compound. A large Ksp means the compound is more soluble.

What happens to the freezing point of a solvent when a solute is added? The boiling point?

The freezing point is lowered because the solute gets in the way of crystal formation. The boiling point is increased due to a lowered vapor pressure. The water molecules are attracted to the solute and are less likely to evaporate causing a lowered vapor pressure. Therefore, more heat must be added to get the vapor pressure to equal the atmospheric pressure.

Would the addition of sodium chloride or the addition of calcium chloride raise the boiling point of water more? Explain.

Calcium chloride would raise the B.P. more because there are more ions released upon dissolving of calcium chloride that sodium chloride.

Is it safer to ice skate on a frozen lake or a frozen oceanic bay in the same conditions? Why?

It would be safer to skate on the lake because the salt dissolved in the oceanic water would depress the freezing point, so that ice would not be as thick.

List 4 factors that effect the dissolving of solutes. What effect do each of these factors have? For which factor(s) would the Ksp change? Would the Ksp go up or down?

Temperature change-heating speeds up molecules so dissolve faster and more. Ksp increases with heat. The opposite is true for gases. Increased temperature causes the dissolved gas to "boil" out.

Stirring-brings solute in contact with solvent so speeds up dissolving. Ksp

unchanged.

Particle Size-more surface area, so more contact of solvent with solute, so speeds up dissolving. Ksp unchanged.

Pressure of gases-pushes molecules closer together so more dissolves. Ksp goes up.

Describe some techniques that could be used to separate the components of a solution.

Your should be able to describe:Distillation (look at your lab)Reverse OsmosisChromatographyDialysis

Problems:What is the molarity of a solution in which 45.0g of sodium nitrate is added to enough water to make 500.0mL of solution?

45.0 g NaNO3 x 1 mole NaNO3 = 0.529 mole NaNO3 85.01 g NaNO3

0.529 mole NaNO3 = 1.06 M .5000 L soln

What is the molality of a solution in which 65.0g of potassium chloride is added to 600.0mL of water?

65.0 g KCl x 1 mole KCl = 0.872 mole KCl74.5 g KCl

0.872 mole KBr = 1.45 m.6000 kg water

How many grams of sodium chloride are needed to make 100.0mL of a 2.0M solution?

.1000 L x 2.0 moles NaCl = 0.20 moles x 58.5 g NaCl = 12g NaCl 1 L solution 1 mole

How many moles of ammonium chloride are used with 2.0L of water

to make a 0.50m solution?

2.0 kg x 0.50 moles = 1.0 moles NH4Cl x 53.5 g NH4Cl = 53.5 g NH4Cl

1 kg 1 mole NH4Cl

What is the expression for the solubility product constant (Ksp) for thedissolving of calcium chloride?

CaCl2 (s) Ca+2 (aq) + 2 Cl-1 (aq)

Ksp + [Ca+2] [Cl-1]2

The solubility of copper(II) hydroxide is 3.4 x 10-7 moles per liter of

solution; what is the solubility product constant (Ksp)?

Cu(OH)2 (s) Cu+2 (aq) + 2 OH-1 (aq) 3.4 x 10-7 0 0 - 3.4 x 10-7 + 3.4 x 10-7 + 6.8 x 10-7 0 3.4 x 10-7 6.8 x 10-7

Ksp = [Cu+2] [OH-1]2 = [3.4 x 10-7] [6.8 x 10-7]2

= 1.6 x 10-19

If 1.2g of SrCrO4 will dissolve in 1 liter of solution, calculate the Ksp.

1.2g SrCrO4 x 1 mole SrCrO4 = 5.9 x 10-3 moles/L = Molarity 203.6g SrCrO4

SrCrO4 (s) Sr+2 (aq) + CrO4-2 (aq)

[5.9 x 10-3] 0 0 - 5.9 x 10-3 + 5.9 x 10-3 + 5.9 x 10-3 0 [5.9 x 10-3] [ 5.9 x 10-3]

Ksp = [Sr+2] [CrO4-2]

= [5.9 x 10-3] [5.9 x 10-3] = 3.5 x 10-5

If 0.85g of MgF2 will dissolve in 0.50 liters of solution, calculate the Ksp.

0.85g MgF2 x 1 mole MgF2 = 0.14 moles 62.27g MgF2

M = 0.14 moles MgF2 = 0.027M 0.50 L solution

MgF2 (s) Mg+2 (aq) + 2 F-1 (aq) [0.027] 0 0 - 0.027 +0.027 + 0.054 0 [0.027] [0.054]

Ksp = [Mg+2] [F-1]2 = [0.027] [0.054]2

= 7.8 x 10-5

Is the SrCrO4 or the MgF2 more soluble? How do you know?The MgF2 is more soluble because it has a higher Ksp.

If the Ksp is 2 x 10-16 at 250, what is the molarity of a saturated solution at this temperature?

PbCrO4 (s) ------ Pb+2 (aq) + CrO4 -2 (aq) 2 x 10-16 = [Pb+2 ] [CrO4-2 ]

x 0 0 - x + x + x 2 x 10-16 = x2

0 x x1.4 x 10-8 M = [PbCrO4 ]

If the Ksp for the dissolving of CoS in water is 4.9 x 10-22, calculate the solubility in moles per liter of solution (maximum molarity).

CoS (s) Co+2 (aq) + S-2 (aq) x 0 0

- x + x + x 0 x x

4.9 x 10-22 = [Co+2] [S-2]4.9 x 10-22 = [x] [x] or x2

x = 2.2 x 10-11 M

The solubility product constant for the dissolving of AgI in water is 8.5 x 10-17 . Calculate the solubility of AgI in grams per one liter ofsolution.

AgI (s) Ag+1 (aq) + I-1 (aq)

x 0 0- x + x + x 0 x x

Ksp = [Ag+1] [I-1]8.5 x 10-17 = [x][x]x = 9.2 x 10-9 M

9.2 x 10-9 moles x 235 g AgI = 2.2 x 10-6 grams in 1L 1 mole AgI

The Ksp for the dissolving of ZnS is 1.3 x 10-22. Would it be possible to dissolve 0.045g in 780mL of water? If so, is the solution saturated or unsaturated?

ZnS (s) Zn+2 (aq) + S-2 (aq) x 0 0- x + x + x 0 x x

Ksp = [Zn+2] [S-2]1.3 x 10-22 = [x][x]x = 1.1 x 10-11 M

1.1 x 10-11 moles x 1 mole ZnS = 1.07 x 10-9 grams in 1L (1000mL) 97.4 g ZnS

1.07 x 10 -9 g ZnS = x g ZnS x = 8.3 x 10-10 grams 1000 ml 780 mL

so….0.045g will not dissolve in 780mL of water.

What is the freezing point of an NaCl solution that contains 21.2 grams solute in 135g water?

21.2g NaCl x 1 mole NaCl = 0.368 moles NaCl 58.5g NaCl

m = 0.368 moles NaCl = 2.72m

0.135 kg H2O

NaCl (s) Na+1 (aq) + Cl-1 (aq) 2.72 0 0- 2.72 + 2.72 + 2.72 0 2.72 2.72 = 5.44m total particle

tf = 1.86oC/m x 5.44m = 10.1oCso freezing point would be – 10.1oC

Which of the following solutions has (a) the higher boiling point and (b) the higher melting point: 0.35m calcium chloride or 0.9m sucrose

CaCl2 (s) Zn+2 (aq) + 2 Cl-1 (aq) 0.35 0 0- 0.35 + 0.35 + 0.70 0 0.35 0.70 = 1.05m total particles

C12H22O11 (s) C12H22O11 (aq) 0.9 0 - 0.9 + 0.9 0 0.9 = 0.9 total particle

therefore CaCl2 causes the larger freezing point depression and the larger boiling point elevation. So, CaCl2 has the higher boiling point and sucrose has the higher melting (freezing) point.

What are the boiling points of the above solutions if the Kb for water is 0.512oC/m?

For CaCl2 tb = 0.512oC/m x 1.05m = 0.54oCSo it boils at 100.54oC.

For Sucrose tb = 0.512oC/m x 0.9m = 0.46oC

So it boils at 100.46oC

Arrange the following solutions in order of decreasing freezing point:0.10m Na3PO4 dissociates to 4 particles so 0.40m total0.35m NaCl dissociates to 2 particles so 0.70m total0.20m MgCl2 dissociates to 3 particles so 0.60m total0.15m C12H22O11 does not dissociate, so 0.15m total

C12H22O11, Na3PO4, MgCl2, NaCl = lowest freezing point

Use the chart on assignments 6 and 7 to answer the following:

Determine if the solutions above are saturated, unsaturated, or supersaturated if prepared at 25oC. (hint: you have to express as grams in 100g water)

9 g of NaNO3 in 100 g of water is unsaturated at any temp.

45g NaNO3 = x g NaNO3 500g H2O 100g H2O

10.8 g of KBr in 100 g of water is unsaturated at any temp.

10.8g KCl = x g KCl 600g H2O 100g H2O

12 g NaCl in 100 g of water is unsaturated at any temp.

2.68 g of NH4Cl in 100 g of water is unsaturated at any temp.

53.5g NH4Cl = x g NH4Cl 2000g H2O 100g H2O

Calculate the Ksp of a saturated solution of potassium nitrate at 25oC.

KNO3 (s) ------- K+ (aq) + NO3 (aq)

Ksp = [K+ ] [NO3]

At 25 degrees, about 40 grams of potassium nitrate will dissolve in 100 g of water:

40 g x 1 mole KNO3 = 0.396 moles KNO3 101 g KNO3

0.396 moles KNO3 = 4.0 M (about) .100 L

so…. Ksp = [4.0 ] [4.0] = 16

How many grams of sodium chloride will dissolve in 100g of water at 30oC?

About 38 g NaCl will dissolve in 100g H2OHow many grams of ammonium chloride will dissolve in 122g of water at 15oC?

35 g NH4Cl = x g NH4Cl = 42.7 g NH4Cl100 g water 122 g water

How many grams will precipitate out of 100mL of a solution of potassium chloride that is saturated at 50oC if the solution is cooled to 10oC?

At 50 degrees, about 40 g will dissolve. At 10 degrees, only about 30 grams will dissolve. About 10 grams will precipitate out.

A solution of NaCl is saturated at 25oC. At what temperature would the solution freeze.

At 25oC, about 38g of NaCl will dissolve in 100g water.

38g NaCl x 1 mole NaCl = 0.65 moles NaCl 58.5g NaCl

m = 0.65 moles NaCl = 6.5m0.100 kg H2O

NaCl (s) Na+1 (aq) + Cl-1 (aq) 6.5 0 0

- 6.5 + 6.5 + 6.5 0 6.5 6.5 = 13m total particle

tf = 1.86oC/m x 13m = 23.9oCso freezing point would be – 23.9oC